1 MW power is to be delivered from a power station to town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if (i) power is transmitted at 220 V . Comment on the feasibility of doing this. (ii) a step-up transformer is used to boost the voltage to 11000 V , power transmitted, then a step-down transformer is used to bring voltage to 220 V . $ ( ρ cu = 1.7 × 1 0 − 8 SI unit ) $

Question

1 MW power is to be delivered from a power station to town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if (i) power is transmitted at 220 V . Comment on the feasibility of doing this. (ii) a step-up transformer is used to boost the voltage to 11000 V , power transmitted, then a step-down transformer is used to bring voltage to 220 V . $ ( ρ cu ​ = 1.7 × 1 0 − 8 SI unit ) $,

MARKS App · Accepted Answer

(i) According to the question, the town is 10 km away, length of pair of copper wires used, $ L = 20 km = 20000 m R es i s t an ceo f \mathrm{Cu} w i res , R = A ρl ​ = π ( r ) 2 ρl ​ A s g i v e n t ha t , ho_{\mathrm{Cu}}=1.7 	imes 10^{-8} ; \mathrm{r}=0.5 \mathrm{~cm} = 3.14 ( 0.5 × 1 0 − 2 ) 2 1.7 × 1 0 − 8 × 20000 ​ = 4Ω A s g i v e n t ha t , \mathrm{P}=1 \mathrm{MW}=10^6 w a ttS o , \mathrm{I} a t 220 \mathrm{~V}, \quad \mathrm{P}=\mathrm{VI}=10^6 \mathrm{~W} , I = 220 1 0 6 ​ = 0.45 × 1 0 4 A P o w er l oss P ​ = RI 2 = 4 × ( 0.45 ) 2 × 1 0 8 W = 8.26 × 1 0 7 W ​ P o w er l oss inh e a t in g as 82.6 MW ​ = 82.6 MW > 1 0 6 W ​ S o , t hi s m e t h o d c ann o t b e u se df or t r an s mi ss i o n . ( ii ) Wh e n p o w er \mathrm{P}=10^6 \mathrm{~W} i s t r an s mi tt e d a t 11000 \mathrm{~V} . V ′ I ′ = 1 0 6 W = 11000 I ′ C u rre n t d r a w n , \quad I^{\prime}=\frac{1}{1.1} 	imes 10^2 	herefore \quad \mathrm{R}_{\mathrm{Cu}}=4 \Omega ( P ) P o w er l oss =\mathrm{RI}^2=\frac{1}{1.21} 	imes 4 	imes 10^4 \mathrm{P}=3.3 	imes 10^4 w a ttS o , F r a c t i o n o f p o w er l oss =\frac{3.3 	imes 10^4}{10^6}=0.033 He n ce , ( P o w er l oss )=3.3 \%$

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